Question: $\sum\limits_{k=1}^{425 }{{(9k -3219)}}=$
What is the question asking for? The question is asking for the sum of the values of $9k -3219$ from $k = 1$ to $k = 425 $ : $(9 \cdot 1 -3219) + (9 \cdot 2 -3219) +... + (9\cdot {425} -3219)$ The series is arithmetic because the formula $9k -3219$ is a linear function of $k$. Formula for arithmetic series The sum $S_n$ of a finite arithmetic series is $S_n = \dfrac {\left(a_1 + a_n \right)}{2} \cdot n$ where $a_1$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. What do we need to use the formula? The number of terms $(n = {425})$ is the upper limit of the sigma notation. We need to find $a_1$ (the first term) and $a_{425}$ (the last term). Step 1: Find $a_1$ and $a_{425}$ (the first and the last term) $a_1 = 9(1) -3219 = {-3210}$ $a_{425} = 9(425) -3219 = {606}$ Step 2: Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac {\left(a_1 + a_n \right)}{2} \cdot n \\\\ S_{{425}}&= \dfrac {\left({-3210} + {606} \right)}{2} \cdot {425} \\\\ S_{{425}} &= -1302 \left(425\right) \\\\ S_{{425}} &= -553{,}350\end{aligned}$ The answer $ -553{,}350 $